a car moving with a constant acceleration covers the distance between two points 60m apart in 6s.?

Question

its velocity as it passes the second point is 15m/s.
a)what was the speed at the first point?
b)what is the constant acceleration?

Answer 1

1) 5m/s
2)5/3 ms^(-2)

Answer 2

Distance lined by applying physique, s = 60 m Time to conceal distance, t = 6 s very final speed, v = 15 m/s Acceleration, a = ? preliminary speed, u = ? s = v*t - 0.5a*t^2 60 =(15)(6) - 0.5 a (6)^2 18a = 30 a = 30/18 = 5 /3 = a million.sixty seven m/s^2 time-honored speed x time = distance lined ( a million/2 ) ( u + v )* t = s u = 2s/t - v = 2*60/6 - 15 = 5 m/s Verification: s = u*t + 0.5a*t^2 = 5 (6) + 0.5 ( 5/3)(6^2) =60 m

Answer 3

Simple really
avg velocity = avg velocity
or
Total dist/ time = init. velocity + final vel./2 (During constant acc)
there fore

10= 15 + x/2
= 20 - 15 = 5m/s
initial velocity = 5m/s
there fore acc
=
v^2 = u^2 +2as
a = 200/120
=
1.67m/s^2

Answer 4

From the general equation
d = Vo*t + 0.5*a*t^2 , or
d = (Vo + Vf) * t /2 where d= distance, Vo = initial velocity, Vf= final velocity, t = time elapsed, a = acceleration.
Vf = Vo + a*t
From those equations you can solve every unknown.

Answer 5

S=60m
u=?
v=15m/s
t=6s
a=?

v-u=at
15-u=a*6
u=15-a*6

S=ut+1/2at*t
60=u*6+1/2*a*6*6
60=(15-6a)*6+1/2*a*6*6
60=90-36a+18a
60=90-18a
18a=30
3a=5
a=4/3
=1.67m/s*s

u=15-6a
=15-6*5/3
=5m

Accleration=1.67
Initial velocity=5

Answer 6

The speed at fisrt point is 5 m/s
The acceleration is 1.67 m/s^2

Answer 7

a) 5m/s
b) 1.6m/s/s